Question

among college students, the proportion p who say theyre interested in their congressional districts election results has traditionally been 75%. after a series of debates on campuses, a political scientist claims that the proportion of college students who say theyre interested in their districts election results is more than 75%. a poll is commissioned, and 225 out of a random sample of 275 college students say theyre interested in their districts election results. is there enough evidence to support the political scientists claim at the 0.05 level of significance? perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) is there enough evidence to support the political scientists claim that the proportion of college students who say theyre interested in their districts election results is more than 75%? yes no

Explanation:

Step1: State the hypotheses

The null hypothesis $H_0$ is the status - quo, and the alternative hypothesis $H_1$ is the claim. The traditional proportion is $p_0 = 0.75$.
$H_0:p = 0.75$
$H_1:p>0.75$

Step2: Determine the test - statistic type

We are testing a hypothesis about a proportion, so we use the z - test for a proportion. The formula for the test - statistic in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$, where $\hat{p}=\frac{x}{n}$, $x$ is the number of successes, and $n$ is the sample size.

Step3: Calculate the sample proportion $\hat{p}$

Given $x = 225$ and $n = 275$, then $\hat{p}=\frac{225}{275}\approx0.818$.

Step4: Calculate the test - statistic $z$

Substitute $\hat{p}=0.818$, $p_0 = 0.75$, and $n = 275$ into the z - formula:
\[

$$\begin{align*} z&=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\\ &=\frac{0.818 - 0.75}{\sqrt{\frac{0.75\times(1 - 0.75)}{275}}}\\ &=\frac{0.068}{\sqrt{\frac{0.75\times0.25}{275}}}\\ &=\frac{0.068}{\sqrt{\frac{0.1875}{275}}}\\ &=\frac{0.068}{\sqrt{0.000681818}}\\ &=\frac{0.068}{0.0261}\\ &\approx2.605 \end{align*}$$

\]

Step5: Calculate the p - value

Since this is a right - tailed test, the p - value is $P(Z>z)$. Using the standard normal distribution table, $P(Z > 2.605)=1 - P(Z\leq2.605)$. From the standard normal table, $P(Z\leq2.605)\approx0.9955$, so the p - value is $1 - 0.9955 = 0.0045$.

Step6: Make a decision

The significance level is $\alpha=0.05$. Since the p - value ($0.0045$) is less than $\alpha = 0.05$, we reject the null hypothesis.

Answer:

(a) $H_0:p = 0.75$, $H_1:p>0.75$
(b) z - test for a proportion
(c) $2.605$
(d) $0.005$
(e) Yes